Talk:Density matrix

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Would be helpful to tie this in to density operator as discusssed in quantum logic. CSTAR 23:03, 19 May 2004 (UTC)

Any objections to moving this page back to "Density matrix" where it belongs? --V79 16:35, 2005 Jun 15 (UTC)

Done --V79 13:08, August 7, 2005 (UTC)

The whole "C*-algebraic formulation of density states" section seems pretty unnecessary to me - right now, it's a lot of jargon that doesn't even make sense to a physicist. Also, a statement like "It is now generally accepted that the description of quantum mechanics in which all self-adjoint operators represent observables is untenable," definitely needs a reference. I will delete the section unless someone objects. --130.126.230.244 17:20, 27 January 2006 (UTC)

Done. If someone feels like rewriting the section to be more explanatory and less jargon-filled, go ahead. 130.126.230.244 16:22, 30 January 2006 (UTC)

The section has been replaced. It could be more explanatory, but as it stands is better than nothing. Those who find it useless may simply ignore it. Archelon 16:58, 5 May 2006 (UTC)

1 question on C* algebra formulation
2 Comment on recent edit
3 merge
4 Entropy
5 Recent edit on equivalence of ensembles
6 Here's an answer
7 Beating the dead horse
8 Spinors from Density Matrices
9 Orders of density matrices

[edit] question on C* algebra formulation

so what's the problem with self adjoint operators as observables? in the finite dim case, the C* algebraic and SA operator formulations are the same. if one is to completely abandon the SA operator formulation, how does one come up with the C*-algebra of observables in the first place? As stated in the article, the GNS lets you recover the Hilbert space, is this not the state space you start with? Mct mht 01:41, 22 May 2006 (UTC)

For one thing, superselection rules. For another, one needs to tie in observables to geometrical structure and symmetry; Geometrical structure as in a "presheaf" kind of association between open sets and observables (in the Streater-Wightman axioms for instance. There's more to it than this of course, but this is the 2 minute summary I could most quickly think of.--CSTAR 02:03, 22 May 2006 (UTC)

here is a more elementary issue, but could also be a problem. as stated in article, if your observables are compact operators, then the states are precisely what's defined in the article. but if you wanna define a quantum operation in the Schrodinger picture, then it's a map between trace-class operators. but the dual map, between observables, is now between the full space of bounded operators. there seems to be some inconsistency there. Mct mht 00:50, 18 June 2006 (UTC)

Ultraweak continuity takes care of this.--CSTAR 02:25, 18 June 2006 (UTC)

so it seems one needs to use the Heisenberg picture map, between observables, in general. it's probably a good idea to modify the article quantum operation a bit accordingly. Mct mht 02:26, 18 June 2006 (UTC)

Yikes No. That article is formulated in terms of the Schrodinger picture. The dual of a quantum operation (on trace class operators) is ultraweakly continuous.--CSTAR 02:36, 18 June 2006 (UTC)

[edit] Comment on recent edit

CSTAR: ok, obviously they're not measures in the measure theoretic sense, but one can see the formal resemblence. and they are called quantum probability measures (was probably your contribution, if i have to guess) in their own right anyway. calling them just that makes the heuristic comment on mixed states as probability distribution on states (common in physics literature) more clear. Mct mht 16:57, 22 May 2006 (UTC)

[edit] merge

the articles pure state and mixed state can easily be merged with this one. Mct mht 05:20, 22 May 2006 (UTC)

Yep.--CSTAR 17:03, 22 May 2006 (UTC)

Support. Nick Mks 15:25, 8 June 2006 (UTC)

[edit] Entropy

I've made some additions to the article consisting of expansion & clarifications of exisiting material and also linking measurement with entropy. See what you think. --Michael C. Price ^talk 20:51, 23 June 2006 (UTC)

[edit] Recent edit on equivalence of ensembles

Quite briefly, I don't believe it. Is there a citation? The assumption is that ensembles are in 1-1 correspondence with operators of the form $A U$ where $A$ is non-negative trace class of trace $1$ and $U$ unitary.

it's not claimed that A is nonnegative. Mct mht 05:15, 29 June 2006 (UTC)

I still don't believe it,

it's also not claimed that A has trace 1. Mct mht 21:00, 29 June 2006 (UTC)

I still don't believe it. --CSTAR 21:07, 29 June 2006 (UTC)

Why? Ensembles have the structure of a convex set. One might naturally associate to an ensemble to a probability measure supported on the convex set of density operators, but from there to the characterization given in the most recent edit, is in my view unjustified. --CSTAR 04:56, 29 June 2006 (UTC)

This might be true for a very special kind of ensemble, in which the components are linearly independent.--CSTAR 05:04, 29 June 2006 (UTC)

let ρ be a mixed state. a square root factorization of ρ is of the from ρ = Σ v_i v_i ^*. from this we see that there is a one-to-one correspondence between such factors and ensembles describing ρ. combine this with the unitary freedom of square roots and that gives what's being claimed. Mct mht 05:11, 29 June 2006 (UTC)

As I said, I don't see what this has to do with ensembles. Please define what you mean by an ensemble. Could you provide a citation?

I think what you are trying to say is that if we have a convex representation

ρ =	∑	λ_iτ_i
	i

where

τ i

are density states, then this is an ensemble (the "this" being the pair consisting of the mixture coefficients and the finite sequence of mixed states τ_i) and the failure of uniqueness is due to the multiplicity of such convex representations.

--CSTAR 14:31, 29 June 2006 (UTC)

yes, that's the, i believe pretty common in physical context, definition of an ensemble. if p_i is a discrete probability distribution, v_i a family of pure states, then {p_i, v_i}_i is called an ensemble. Mct mht 21:11, 29 June 2006 (UTC)

Well, then more generally we can write

ρ =	∫	τdμ(τ)
	S

where μ is a probability measure on the compact convex space of states. We can regard the measure μ as an ensemble representing the state τ. The non-uniqueness follows from the failure of unique representation by probability measures, even probability measures which are supported on the set of extreme points. --CSTAR 21:21, 29 June 2006 (UTC)

yes, agreed. didn't you object to calling those things probability measures. :-) Mct mht 21:38, 29 June 2006 (UTC)

No. If I recall correctly, what I objected to here was calling a completely additive function on the orthocomplemented lattice of projections on a Hilbert space a probability measure. There is an affine bijection between such functions and density operators. A probability mesure on states as mentioned in my remark above (in this talk page) is a probability measure in the classical sense, that is, a set function on the σ-algebra of Borel sets.--CSTAR 21:48, 29 June 2006 (UTC)

In conclusion, then the non-uniqueness follows from the multiplicity of representations of the form

ρ =	∫	τdμ(τ)
	S

by probability measures μ supported on the compact convex set of states S. This can be described as non-uniqueness of convex representations. What does this have to do with non-uniquess of factorizations (except in very special cases?) --CSTAR 00:45, 30 June 2006 (UTC)

in the formulation of my comments above, this is precisely the non uniqueness of square root factorization of positive semidefinite operators restated in this context. take for instance the finite dim case. given a mixed state described by the ensemble {v_i}, where the probabilities are absorbed into the states, for notational convenience. so ρ = Σ v_i v_i ^*. another ensemble {w_i} describes the same state iff there exists a unitary U such that [v_i] U = [w_i]. Mct mht 01:39, 30 June 2006 (UTC)

This would imply that all convex representations of a state have the same cardinality.

that's not true. let me clarify. we don't assume [v_i] is full rank. same goes for [w_i]. when an ensemble v₁...v_m is such that m ≤ n, we simply append columns of zeros to get a square matrix [v_i 0]. as a (somewhat common) abuse of notation from linear algebra, above i have put [v_i]. all i am doing is stating a fact from linear algebra and relate to this context. Mct mht 03:13, 30 June 2006 (UTC)

Could you provide a citation for this fact?--CSTAR 03:17, 30 June 2006 (UTC)

PS BTW, in the example I suggested, no matter how much appending you do, that trick isn't going to work.--CSTAR 03:19, 30 June 2006 (UTC)

in all my comments, ensemble elements are pure states (vectors). i thought that was clear. that's not so in the e.g. you gave. Mct mht 03:30, 30 June 2006 (UTC)

But this isn't true. To take a trivial example; any state ρ is representable as a trivial convex combination of 1 times itself ρ= 1 × ρ (cardinbality 1) and if it's a non-extremal state, at least one other non-trivial convex combination of two other states (cardinality two).--CSTAR 02:20, 30 June 2006 (UTC)

BTW, in general the set of states ain't compact, no? Mct mht 03:14, 30 June 2006 (UTC)

appended after CTAR's reply below: i meant norm compact above. Mct mht 03:23, 30 June 2006 (UTC)

In the finite dimensional case, all TVS topologies are the same.--CSTAR 03:24, 30 June 2006 (UTC)

Yes, it's always compact in w* topology (even for C*-algebras) That's a result of fundamnetal importance. This follows from the Banach-ALouglu theorem.--CSTAR 03:17, 30 June 2006 (UTC)

Re:In all my comments, ensemble elements are pure states (vectors). i thought that was clear You thought that was clear? Um, it was not clear to me. Now with that caveat, the result you are claiming is equivalent to the following: Representations of the form

τ =	∑	λ_iE_i
	i

where the $E i$ are selfadjoint projections (that are however not assumed to be pairwise orthogonal) are uniquely determined up to unitaries. That is if

τ =	∑	λ_iF_i
	i

then there is a unitary U such that

E i = U F i U *

Frankly, I don't believe this is true. (If the projections are assumed pairwise orthogonal, of coure it's just the spectral theorem). This claim would imply that all probability measures on the compact convex set of states, supported on the extreme points, are "unitarily equivalent" provided they have the same center of mass. I would be surprised f that were true in anything but the abelian case. However, if you provide me with a reference, I'll believe it.--CSTAR 03:49, 30 June 2006 (UTC)

CSTAR, that is not at all equivalent to what was claimed! Mct mht 04:05, 30 June 2006 (UTC)

If a states is an extreme point, it is a rank 1 projection. therefore, if

τ =	∑	λ_iA_i
	i

where the A_i are extreme points, then the A_i must be rank 1 projections.

Now I thought you just claimed that two representations as convex combinations of extreme poinys must be equivalent "unitarily"; What else could you possibly mean other than what I wrote? --CSTAR 04:15, 30 June 2006 (UTC)

Anyway, please provide a citation with whatever it is you are claiming. That will save everybody a lot of time.--CSTAR 04:16, 30 June 2006 (UTC)

CSTAR, i don't know, man, it's probably in most linear algebra books. It's pretty standard. For instance, Choi used this to show how the Kraus operators of a CP map are related by a unitary matrix. Again, the claim is simply the following: Let A = M M^* be a positive semidefinite matrix. Then N satisfies A = N N^* iff N = M U for some unitary U, where all matrices are square (If we drop the square assumption, then U need not be square, but it is isometric). Mct mht 04:53, 30 June 2006 (UTC)

OK that I believe and is easy to prove. What that has to do with representations of ensembles however, is not at all clear.--CSTAR 05:06, 30 June 2006 (UTC)

it is related exactly in the sense i describe before. if A is a state, the column vectors of M gives an ensemble describing the state, same goes for N (the probabilities can be recovered easily). if you take the unique positive square root, you get an ensemble of orthonormal states ... Mct mht 05:19, 30 June 2006 (UTC)

I guess we're back to square 1. My original question above remains as far as I can see, unanswered: What is the relation between "square root factorizations" and ensembles? Do you have a citation for this relation?

Re: the column vectors of M gives an ensemble describing the state, I'm assuming here A = M*M. What's the ensemble (understood as some convex combination of pure states) that gives the state A? Presumably, they're related to the columns or rows of M, but I'm sorry I don't see this.

Re: if you take the unique positive square root, you get an ensemble of orthonormal states ...

In fact, if you take the unique positive square root, you get a positive operator. The columns of this operator are not going to be orthogonal (except in some trivial cases). How do you get an ensemble of orthonormal states. Do you mean the spectral decomposition of A? Could you please be more specific, saying how you do this?

reply: say A has spectral resolution A = U Σ U^* = U Σ ^½ Σ ^½ U^*. Let M = U Σ ^½, then it's trivial to verify what i said. Mct mht 02:05, 1 July 2006 (UTC)

Just give me a citation (with the statement of the result about the relation between ensembles and factorizations) and I'll go away.--CSTAR 13:43, 30 June 2006 (UTC)

This applies to couple other related objects as well. Besides Kraus operators, purifications of a given mixed state are related in a similar way. Mct mht 05:31, 30 June 2006 (UTC)

What's the "this" that applies to other related objects?--CSTAR 13:43, 30 June 2006 (UTC)

CSTAR, i am done with this particular discussion. one can't get much more explicit than the explanations i've given. i am sorry you're not convinced. this is completely elementary and standard. Mct mht 02:05, 1 July 2006 (UTC)

I'm sorry I still don't believe what you said. In fact, to be honest, I don't understand what you said about the relation between ensembles and factorizations. I can't believe that it's standard; if indeed it were standard it would be easy to come up with a source in the form page X of Y which says Z. But, suit yourself. --CSTAR 04:03, 1 July 2006 (UTC)

PS If somewhat has understood mct mht's point about the relation between ensembles and factorizations and I'm just being dense, I'll listen. --CSTAR 04:07, 1 July 2006 (UTC)

[edit] Here's an answer

First of all definition of an ensemble: this is an indexed family of pairs

$p_i, \mid \psi_i\rangle$

with

$p_i \geq 0, \quad \sum_i p_i =1$

and

$\mid \psi_i\rangle \,$

a unit ket vector. The corresponding density matrix is the convex combination of rank one projections

$\sum_i p_i \mid \psi_i\rangle \langle \psi_i \mid$

As Nielsen and Chuang point out, it is actually more convenient to absorb the probability coefficient into the state by replacing thekets with "renormalized" kets:

$\mid \psi_i \rangle \,$ by $\mid \sqrt{p_i} \psi_i \rangle$

Thus with this renormalized formulation, the corresponding density matrix is

$\sum_i \mid \psi_i\rangle \langle \psi_i \mid.$

Theorem. Two ensembles ψ, ψ' define the same density state iff there is a unitary matrix

u i j

such that

$\mid \psi_i\rangle = \sum_{j} u_{ij} \mid \psi'_j\rangle$

This is Theorem 2.6 of Nielsen and Chuang. --CSTAR 04:43, 1 July 2006 (UTC)

OK, that seems clear. I'd rather not work with renormalised states, though, and re-express your final equation as:

$\mid \psi_i\rangle \sqrt p_i = \sum_{j} u_{ij} \mid \psi'_j\rangle \sqrt {p_j^'}$

--Michael C. Price ^talk 05:18, 1 July 2006 (UTC)

Are you going to replace the MM* stuff (which completely lost me) with the above crystal-clear construction?

I have another query; this line in the introduction seems false:

The description of states in Dirac notation by ket vectors is not sufficient to describe the effect of quantum operations, such as measurement, on a quantum mechanical ensemble.

Any mixture of kets can be analysed by treating each ket separately. I suggest deletion. The point (about the merit of density matrices in statistical analysis) seems adequately made by the surrounding text.--Michael C. Price ^talk 06:03, 1 July 2006 (UTC)

I trust your judgement on replacements and deletions (and yes I prefer your non-normalized form of Nielsen and Chuang Theorem 2.6.) I feel I've done enough damage for one day.--CSTAR 06:10, 1 July 2006 (UTC)

[edit] Beating the dead horse

OK it may be standard, but here is the the explicit relationship:

By definition, a positive rank one operator on H is one of the form

$\mid \phi\rangle \langle \phi \mid$

Recall that an ensemble representing a density matrix is a sequence of positive rank one operators which add up to A.

Fix an orthonormal basis for H. The matrix of such an operator is

$\langle e_i \mid \phi\rangle \langle \phi \mid e_j \rangle = \phi_i \overline{\phi_j}$

Conversely any operator whose matrix has that form is rank 1. Now suppose

A = ΨΨ *

Thus

$A_{ij}= \sum_k \psi_{ik} \overline{\psi_{jk}}$

Consider the sequence of rank one operators with matrices

$\Phi^k_{i j} = \psi_{ik} \overline{\psi_{jk}}$

Then clearly

$A_{ij}= \sum_k \Phi_{ij}^k$

This is an ensemble representing A. Conversely revcersing the argument, one sees that any ensemble representing A has this form. --CSTAR 21:21, 1 July 2006 (UTC)

[edit] Spinors from Density Matrices

The page includes the usual method of obtaining the density matrix from spinors but does not show the less well known method of obtaining spinors from density matrices. I'll go ahead and add it in. If you haven't seen the method, see Julian Schwinger's "Quantum Kinematics and Dynamics" or www.DensityMatrix.com .CarlAB 01:16, 6 October 2006 (UTC)

[edit] Orders of density matrices

So far this article has ignored the idea of various orders of density matrices. Also, some of the particular expressions are in fact for reduced density operators where the true expressions for the reduced density *matrices* is not given.

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